Problem B Boxes in a Line
手写链表。。其实很简单的。。。。
注:对相邻的元素需要特判。。。。。
Problem B
Boxes in a Line
You have n boxes in a line on the table numbered 1~n from left to right. Your task is to simulate 4 kinds of commands:
- 1 X Y: move box X to the left to Y (ignore this if X is already the left of Y)
- 2 X Y: move box X to the right to Y (ignore this if X is already the right of Y)
- 3 X Y: swap box X and Y
- 4: reverse the whole line.
Commands are guaranteed to be valid, i.e. X will be not equal to Y.
For example, if n=6, after executing 1 1 4, the line becomes 2 3 1 4 5 6. Then after executing 2 3 5, the line becomes 2 1 4 5 3 6. Then after executing 3 1 6, the line becomes 2 6 4 5 3 1. Then after executing 4, then line becomes 1 3 5 4 6 2
Input
There will be at most 10 test cases. Each test case begins with a line containing 2 integers n, m(1<=n, m<=100,000). Each of the following m lines contain a command.
Output
For each test case, print the sum of numbers at odd-indexed positions. Positions are numbered 1 to n from left to right.
Sample Input
6 4
1 1 4
2 3 5
3 1 6
4
6 3
1 1 4
2 3 5
3 1 6
100000 1
4
Output for the Sample Input
Case 1: 12
Case 2: 9
Case 3: 2500050000
The Ninth Hunan Collegiate Programming Contest (2013)
Problemsetter: Rujia Liu
Special Thanks: Feng Chen, Md. Mahbubul Hasan
#include <iostream>
#include <cstdio>
#include <cstring> #define toleft p
#define toright p^1 using namespace std; const int INF=0x3f3f3f3f;
typedef long long int LL; int lian[][],n,m,p; void Init()
{
p=;
for(int i=;i<=n;i++)
{
lian[i][]=i-;
lian[i][]=i+;
}
lian[][toleft]=INF;
lian[][toright]=;
lian[n+][toleft]=n;
lian[n+][toright]=INF;
} void cmdfour()
{
p=p^;
} void cmdone(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xr==y) return; lian[xl][toright]=xr;
lian[xr][toleft]=xl; lian[x][toright]=y;
lian[x][toleft]=yl; lian[yl][toright]=x;
lian[y][toleft]=x;
} void cmdtwo(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xl==y) return; lian[xl][toright]=xr;
lian[xr][toleft]=xl; lian[x][toleft]=y;
lian[x][toright]=yr; lian[y][toright]=x;
lian[yr][toleft]=x;
} void cmdthree(int x,int y)
{
int xl=lian[x][toleft],xr=lian[x][toright];
int yl=lian[y][toleft],yr=lian[y][toright]; if(xr!=y&&xl!=y)
{
lian[x][toleft]=yl;
lian[x][toright]=yr;
lian[yl][toright]=x;
lian[yr][toleft]=x; lian[y][toleft]=xl;
lian[y][toright]=xr;
lian[xl][toright]=y;
lian[xr][toleft]=y;
}
else
{
if(xr==y)
{
lian[x][toright]=yr;
lian[x][toleft]=y; lian[y][toright]=x;
lian[y][toleft]=xl; lian[xl][toright]=y;
lian[yr][toleft]=x;
}
else if(xl==y)
{
lian[x][toleft]=yl;
lian[x][toright]=y; lian[y][toleft]=x;
lian[y][toright]=xr; lian[xr][toleft]=y;
lian[yl][toright]=x;
} }
} LL getSum()
{
LL ans=;
int cnt=,pos=;
if(n%==||p==)
{
if(p) p=;
while(true)
{
if(cnt&) ans+=(LL)lian[pos][toright];
pos=lian[pos][toright];
cnt++;
if(cnt>=n+) break;
}
}
else if(p==)
{
if(p) p=;
while(true)
{
if(cnt%==) ans+=(LL)lian[pos][toright];
pos=lian[pos][toright];
cnt++;
if(cnt>=n+) break;
}
}
return ans;
} int main()
{
int cas=,cmd,x,y;
while(scanf("%d%d",&n,&m)!=EOF)
{
Init(); while(m--)
{
scanf("%d",&cmd);
if(cmd==)
{
scanf("%d%d",&x,&y);
cmdone(x,y);
}
else if(cmd==)
{
scanf("%d%d",&x,&y);
cmdtwo(x,y);
}
else if(cmd==)
{
scanf("%d%d",&x,&y);
cmdthree(x,y);
}
else if(cmd==)
{
cmdfour();
}
}
LL ans=getSum();
printf("Case %d: %lld\n",cas++,ans);
}
return ;
}
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